Complete the square to solve for $x$. $x^{2}-15x+56 = 0$
Solution: Move the constant term to the right side of the equation. $x^2 - 15x = -56$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $-15$ , so half of it would be $-\dfrac{15}{2}$ , and squaring it gives us ${\dfrac{225}{4}}$ $x^2 - 15x { + \dfrac{225}{4}} = -56 { + \dfrac{225}{4}}$ We can now rewrite the left side of the equation as a squared term. $( x - \dfrac{15}{2} )^2 = \dfrac{1}{4}$ Take the square root of both sides. $x - \dfrac{15}{2} = \pm\dfrac{1}{2}$ Isolate $x$ to find the solution(s). $x = \dfrac{15}{2}\pm\dfrac{1}{2}$ The solutions are: $x = 8 \text{ or } x = 7$ We already found the completed square: $( x - \dfrac{15}{2} )^2 = \dfrac{1}{4}$